thermo

S K Mondal’s

Thermodynamics

GATE, IES & IAS 20 Years Question Answers

Contents

Chapter – 1: Basic Concepts

Chapter - 2 : First Law of Thermodynamics

Chapter - 3 : Second Law of Thermodynamics

Chapter - 4 : Entropy

Chapter - 5 : Availability, Irreversibility

Chapter - 6 : Thermodynamic Relations

Chapter - 7 : Pure Substances

Chapter - 8 : Properties of Gasses & Gas Mix

Er. S K Mondal

IES Officer (Railway), GATE topper, NTPC ET-2003 batch, 12 years teaching

experienced, Author of Hydro Power Familiarization (NTPC Ltd)

Note

If you think there should be a change in

option, don’t change it by yourself send me a

mail

at

[email protected]

I will send you complete explanation.

Copyright © 2007 S K Mondal

Every effort has been made to see that there are no errors (typographical or otherwise) in the material presented. However, it is still possible that there are a few errors (serious or otherwise). I would be thankful to the readers if they are brought to my attention at the following e-mail address: [email protected]

Basic Concepts

S K Mondal’s

Chapter 1

1.

Basic Concepts

A

SKED

O

BJECTIVE

Q

UESTIONS

(GATE, IES, IAS)

Previous 20-Years GATE Questions

GATE-1. List-I List II [GATE-1998]

A. Heat to work 1. Nozzle

B. Heat to lift weight 2. Endothermic chemical reaction

C. Heat to strain energy 3. Heat engine

D. Heat to electromagnetic energy 4. Hot air balloon/evaporation

5. Thermal radiation

6. Bimetallic strips

Codes: A B C D A B C D

(a) 3 4 6 5 (b) 3 4 5 6

(c) 3 6 4 2 (d) 1 2 3 4

Open and Closed systems

GATE-2. An isolated thermodynamic system executes a process, choose the correct

statement(s) form the following [GATE-1999]

(a) No heat is transferred (b) No work is done

(c) No mass flows across the boundary of the system (d) No chemical reaction takes place within the system

Quasi-Static Process

GATE-3. A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m3_{. It expands quasi-statically at constant temperature to a final volume}

of 0.030 m3_{. The work output (in kJ/kg) during this process will be: [GATE-2009]}

(a) 8.32 (b) 12.00 (c) 554.67 (d) 8320.00

Free Expansion with Zero Work Transfer

GATE-4. A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is TRUE at the end of above process?

(a) The internal energy of the gas decreases from its initial value, but the enthalpy

remains constant [GATE-2008]

Basic Concepts

S K Mondal’s

Chapter 1

(c) Both internal energy and enthalpy of the gas remain constant (d) Both internal energy and enthalpy of the gas increase

GATE-5. Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The Work done in the process is given by:

[GATE-1996, IAS-2000]

(a) –∫Pdv (b) +∫Pdv (c) –∫vdp (d) +∫vdp

pdV-work or Displacement Work

GATE-6. In a steady state steady flow process taking place in a device with a single inlet and a single outlet, the work done per unit mass flow rate is given by

outlet inlet

vdp

ω

= −

_∫

, where v is the specific volume and p is the pressure. The

expression for w given above: [GATE-2008]

(a) Is valid only if the process is both reversible and adiabatic (b) Is valid only if the process is both reversible and isothermal (c) Is valid for any reversible process

(d) Is incorrect; it must be

outlet inlet

vdp

ω

= −

_∫

GATE-7. A gas expands in a frictionless piston-cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01 m3_{. The maximum amount of work}

that could be utilized from the above process is: [GATE-2008]

(a) 0kJ (b) 1kJ (c) 2kJ (d) 3kJ

GATE-8. For reversible adiabatic compression in a steady flow process, the work

transfer per unit mass is: [GATE-1996]

( )

a

∫

pdv

( )

b

∫

vdp

( )

c

∫

Tds

( )

d

∫

sdT

Previous 20-Years IES Questions

IES-1. Which of the following are intensive properties? [IES-2005]

1. Kinetic Energy 2. Specific Enthalpy

3. Pressure 4. Entropy

Select the correct answer using the code given below:

(a) 1 and 3 (b) 2 and 3 (c) 1, 3 and 4 (d) 2 and 4

IES-2. Consider the following properties: [IES-2009]

1. Temperature 2. Viscosity

3. Specific entropy 4. Thermal conductivity

Which of the above properties of a system is/are intensive?

Basic Concepts

S K Mondal’s

Chapter 1

IES-3. Which one of the following is the extensive property of a thermodynamic

system? [IES-1999]

(a) Volume (b) Pressure (c) Temperature (d) Density

IES-4. Consider the following properties: [IES-2009]

1. Entropy 2. Viscosity

3. Temperature 4. Specific heat at constant volume

Which of the above properties of a system is/are extensive?

(a) 1 only (b) 1 and 2 only (c) 2, 3 and 4 (d) 1, 2 and 4

Thermodynamic System and Control Volume

IES-5. Assertion (A): A thermodynamic system may be considered as a quantity of

working substance with which interactions of heat and work are studied. Reason (R): Energy in the form of work and heat are mutually convertible.

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A

(c) A is true but R is false [IES-2000]

(d) A is false but R is true

Open and Closed systems

IES-6. A closed thermodynamic system is one in which [IES-1999]

(a) There is no energy or mass transfer across the boundary (b) There is no mass transfer, but energy transfer exists (c) There is no energy transfer, but mass transfer exists

(d) Both energy and mass transfer take place across the boundary, but the mass transfer is controlled by valves

IES-7. Which of the following are intensive properties? [IES-2007]

1. Kinetic energy 2. Thermal conductivity

3. Pressure 4. Entropy

Select the correct answer using the code given below:

(a) 1 and 2 (b) 2 and 3 only (c) 2, 3 and 4 (d) 1, 3 and 4

IES-8. Which of the following is/are reversible process(es)? [IES-2005]

1. Isentropic expansion

2. Slow heating of water from a hot source

3. Constant pressure heating of an ideal gas from a constant temperature source

4. Evaporation of a liquid at constant temperature Select the correct answer using the code given below:

(a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 4

IES-9. Assertion (A): In thermodynamic analysis, the concept of reversibility is that, a

reversible process is the most efficient process. [IES-2001]

Reason (R): The energy transfer as heat and work during the forward process as always identically equal to the energy transfer is heat and work during the reversal or the process.

Basic Concepts

S K Mondal’s

Chapter 1

(c) A is true but R is false (d) A is false but R is true

IES-10. Ice kept in a well insulated thermo flask is an example of which system?

(a) Closed system (b) Isolated systems [IES-2009]

(c) Open system (d) Non-flow adiabatic system

Zeroth Law of Thermodynamics

IES-11. Measurement of temperature is based on which law of thermodynamics?

[IES-2009]

(a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics

IES-12. Consider the following statements: [IES-2003]

1. Zeroth law of thermodynamics is related to temperature 2. Entropy is related to first law of thermodynamics

3. Internal energy of an ideal gas is a function of temperature and pressure 4. Van der Waals' equation is related to an ideal gas

Which of the above statements is/are correct?

(a) 1 only (b) 2, 3 and 4 (c) 1 and 3 (d) 2 and 4

IES-13. Zeroth Law of thermodynamics states that [IES-1996]

(a) Two thermodynamic systems are always in thermal equilibrium with each other. (b) If two systems are in thermal equilibrium, then the third system will also be in

thermal equilibrium with each other.

(c) Two systems not in thermal equilibrium with a third system are also not in thermal equilibrium with each other.

(d) When two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

International Temperature Scale

IES-14. Which one of the following correctly defines 1 K, as per the internationally

accepted definition of temperature scale? [IES-2004]

(a) 1/100th _{of the difference between normal boiling point and normal freezing point of}

water

(b) 1/273.15th _{of the normal freezing point of water}

(c) 100 times the difference between the triple point of water and the normal freezing point of water

(d) 1/273.15th _{of the triple point of water}

IES-15. In a new temperature scale say °ρ, the boiling and freezing points of water at

one atmosphere are 100°ρ and 300°ρ respectively. Correlate this scale with the

Centigrade scale. The reading of 0°ρ on the Centigrade scale is: [IES-2001]

Basic Concepts

S K Mondal’s

Chapter 1

IES-16. Assertion (a): If an alcohol and a mercury thermometer read exactly 0°C at the

ice point and 100°C at the steam point and the distance between the two points is divided into 100 equal parts in both thermometers, the two thermometers

will give exactly the same reading at 50°C. [IES-1995]

Reason (R): Temperature scales are arbitrary.

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false

(d) A is false but R is true

IES-17. Match List-I (Type of Thermometer) with List-II (Thermometric Property) and

select the correct answer using the code given below the [IES 2007]

List-I List-II

A. Mercury-in-glass 1. Pressure

B. Thermocouple 2. Electrical resistant

C. Thermistor 3. Volume

D. Constant volume gas 4. Induced electric voltage

Codes: A B C D A B C D

(a) 1 4 2 3 (b) 3 2 4 1

(c) 1 2 4 3 (d) 3 4 2 1

IES-18. Pressure reaches a value of absolute zero [IES-2002]

(a) At a temperature of – 273 K

(b) Under vacuum condition (c) At the earth's centre

(d) When molecular momentum of system becomes zero

IES-19. The time constant of a thermocouple is the time taken to attain:

(a) The final value to he measured [IES-1997]

(b) 50% of the value of the initial temperature difference (c) 63.2% of the value of the initial temperature difference (d) 98.8% of the value of the initial temperature difference

Work a Path Function

IES-20. Assertion (A): Thermodynamic work is path-dependent except for an adiabatic

process. [IES-2005] Reason(R): It is always possible to take a system from a given initial state to

any final state by performing adiabatic work only.

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false

(d) A is false but R is true

Free Expansion with Zero Work Transfer

IES-21. Match items in List-I (Process) with those in List-II (Characteristic) and select the correct answer using the codes given below the lists:

Basic Concepts

S K Mondal’s

Chapter 1

A. Throttling process 1. No work done

B. Isentropic process 2. No change in entropy

C. Free expansion 3. Constant internal energy

D. Isothermal process 4. Constant enthalpy

Codes: A B C D A B C D

(a) 4 2 1 3 (b) 1 2 4 3 (c) 4 3 1 2 (d) 1 3 4 2

IES-22. The heat transfer, Q, the work done W and the change in internal energy U are

all zero in the case of [IES-1996]

(a) A rigid vessel containing steam at 150°C left in the atmosphere which is at 25°C. (b) 1 kg of gas contained in an insulated cylinder expanding as the piston moves slowly

outwards.

(c) A rigid vessel containing ammonia gas connected through a valve to an evacuated rigid vessel, the vessel, the valve and the connecting pipes being well insulated and the valve being opened and after a time, conditions through the two vessels becoming uniform.

(d) 1 kg of air flowing adiabatically from the atmosphere into a previously evacuated bottle.

pdV-work or Displacement Work

IES-23. One kg of ice at 0°C is completely melted into water at 0°C at 1 bar pressure.

The latent heat of fusion of water is 333 kJ/kg and the densities of water and

ice at 0°C are 999.0 kg/m3 _{and 916.0 kg/m}3, _{respectively. What are the}

approximate values of the work done and energy transferred as heat for the

process, respectively? [IES-2007]

(a) –9.4 J and 333.0 kJ (b) 9.4 J and 333.0 kJ (c) –333.0 kJ and –9.4 J (d) None of the above

IES-24. Which one of the following is the

correct sequence of the three processes A, B and C in the increasing order of the amount of work done by a gas following ideal-gas expansions by these processes?

[IES-2006]

Basic Concepts

S K Mondal’s

Chapter 1

IES-25. An ideal gas undergoes an isothermal expansion from state R to state S in a turbine as shown in the diagram given below:

The area of shaded region is 1000 Nm. What is the amount is turbine work done during the process?

(a) 14,000 Nm (b) 12,000 Nm

(c) 11,000 Nm (d) 10,000 Nm _[IES-2004]

IES-26. Assertion (A): The area 'under' curve on pv plane,

∫

pdvrepresents the work of

reversible non-flow process. [IES-1992]

Reason (R): The area 'under' the curve T–s plane

∫

Tds represents heat of any reversible process.

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false

(d) A is false but R is true

IES-27. If

∫

pdv and −

∫

vdp for a thermodynamic system of an Ideal gas on valuation

give same quantity (positive/negative) during a process, then the process

undergone by the system is: [IES-2003]

(a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal

IES-28. Which one of the following expresses the reversible work done by the system

(steady flow) between states 1 and 2? [IES-2008]

2 2 2 2

1 1 1 1

(a) (b) (c) (d)

∫

pdv −

∫

vdp −

∫

pdv

∫

vdp

Heat Transfer-A Path Function

IES-29. Assertion (A): The change in heat and work cannot be expressed as difference

between the end states. [IES-1999]

Reason (R): Heat and work are both exact differentials.

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false

Basic Concepts

S K Mondal’s

Chapter 1

Previous 20-Years IAS Questions

Thermodynamic System and Control Volume

IAS-1. The following are examples of some intensive and extensive properties:

1. Pressure 2. Temperature [IAS-1995]

3. Volume 4. Velocity

5. Electric charge 6. Magnetisation

7. Viscosity 8. Potential energy

Which one of the following sets gives the correct combination of intensive and extensive properties? Intensive Extensive (a) 1, 2, 3, 4 5, 6, 7, 8 (b) 1, 3, 5, 7 2, 4, 6, 8 (c) 1, 2, 4, 7 3, 5, 6, 8 (d) 2, 3, 6, 8 1, 4, 5, 7

Zeroth Law of Thermodynamics

IAS-2. Match List-I with List-II and select the correct answer using the codes given

below the lists: [IAS-2004]

List-I List-II

A. Reversible cycle 1. Measurement of temperature

B. Mechanical work 2. Clapeyron equation

C. Zeroth Law 3. Clausius Theorem

D. Heat 4. High grade energy

5. 3rd _{law of thermodynamics}

6. Inexact differential

Codes: A B C D A B C D

(a) 3 4 1 6 (b) 2 6 1 3 (c) 3 1 5 6 (d) 1 4 5 2

IAS-3. Match List-I with List-II and select the correct answer: [IAS-2000]

List-I List-II

A. The entropy of a pure crystalline 1. First law of thermodynamics

substance is zero at absolute zero

temperature

B. Spontaneous processes occur 2. Second law of thermodynamics

in a certain direction

C. If two bodies are in thermal 3. Third law of thermodynamics

equilibrium with a third body, then they are also in thermal equilibrium with each other

D. The law of conservation of energy 4. Zeroth law of thermodynamics.

Codes: A B C D A B C D

(a) 2 3 4 1 (b) 3 2 1 4 (c) 3 2 4 1 (d) 2 3 1 4

Basic Concepts

S K Mondal’s

Chapter 1

International Temperature Scale

IAS-4. A new temperature scale in degrees N is to be defined. The boiling and freezing on this scale are 400°N and 100°N respectively. What will be the reading on new scale corresponding to 60°C? [IAS-1995]

(a) 120°N (b) 180°N (c) 220°N (d) 280°N

Free Expansion with Zero Work Transfer

IAS-5. In free expansion of a gas between two equilibrium states, the work transfer

involved [IAS-2001]

(a) Can be calculated by joining the two states on p-v coordinates by any path and estimating the area below

(b) Can be calculated by joining the two states by a quasi-static path and then finding the area below

(c) Is zero

(d) Is equal to heat generated by friction during expansion.

IAS-6. Work done in a free expansion process is: [IAS-2002]

(a) Positive (b) Negative (c) Zero (d) Maximum

IAS-7. In the temperature-entropy diagram

of a vapour shown in the given figure, the thermodynamic process shown by the dotted line AB represents

(a) Hyperbolic expansion (b) Free expansion

(c) Constant volume expansion (d) Polytropic expansion

[IAS-1995]

IAS-8. If

∫

pdv and −

∫

vdp for a thermodynamic system of an Ideal gas on valuation

give same quantity (positive/negative) during a process, then the process

undergone by the system is: [IAS-1997, IES-2003]

(a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal

IAS-9. For the expression

∫

pdv to represent the work, which of the following

conditions should apply? [IAS-2002]

(a) The system is closed one and process takes place in non-flow system (b) The process is non-quasi static

(c) The boundary of the system should not move in order that work may be transferred (d) If the system is open one, it should be non-reversible

IAS-10. Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The Work done in the process is given by:

Basic Concepts

S K Mondal’s

Chapter 1

IAS-11. Match List-I with List-II and select the correct answer using the codes given

below the lists: [IAS-2004]

List-I List-II

A. Bottle filling of gas 1. Absolute Zero Temperature

B. Nernst simon Statement 2. Variable flow

C. Joule Thomson Effect 3. Quasi-Static Path

D. ∫pdv 4. Isentropic Process

5. Dissipative Effect 6. Low grade energy

7. Process and temperature during phase

change.

Codes: A B C D A B C D

(a) 6 5 4 3 (b) 2 1 4 3

(c) 2 5 7 4 (d) 6 1 7 4

pdV-work or Displacement Work

IAS-13. Thermodynamic work is the product of [IAS-1998]

(a) Two intensive properties (b) Two extensive properties

(c) An intensive property and change in an extensive property (d) An extensive property and change in an intensive property

Heat Transfer-A Path Function

IAS-14. Match List-I (Parameter) with List-II (Property) and select the correct answer using the codes given below the lists:

List-I List-II [IAS-1999]

A. Volume 1. Path function

B. Density 2. Intensive property

C. Pressure 3. Extensive property

D. Work 4. Point function

Codes: A B C D A B C D

(a) 3 2 4 1 (b) 3 2 1 4 (c) 2 3 4 1 (d) 2 3 1 4

Basic Concepts

S K Mondal’s

Chapter 1

Answers with Explanation (Objective)

Previous 20-Years GATE Answers

GATE-1. Ans. (a)

GATE-2. Ans. (a, b, c) For an isolated system no mass and energy transfer through the system.

0, 0, 0 or Constant

dQ= dW = ∴ dE = E=

GATE-3. Ans. (a) Iso-thermal work done (W) = 2

1 1 ln V RT V ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 2 1 1 1 ln 0.030 800 0.015 ln 0.015 8.32kJ/kg V P V V ⎛ ⎞ = _{⎜ ⎟} ⎝ ⎠ ⎛ ⎞ = × × _{⎜ ⎟} ⎝ ⎠ =

GATE-4. Ans. (c) It is free expansion. Since vacuum does not offer any resistance, there is no work

transfer involved in free expansion. Here,

2 1

0

δω

=

∫

and Q1-2=0 therefore Q1-2 =

Δ

U

+ W1-2 so,

Δ

U

= 0

GATE-5. Ans. (c) For closed system W =

+

∫

pdv

, for steady flow W =

−

∫

vdp

GATE-6. (c)

GATE-7. Ans. (b) W = Resistance pressure.

Δ

V = 1 ×

Δ

V = 100 × 0.1 kJ = 1kJ

GATE-8. Ans. (b) W = −

_∫

vdp

Previous 20-Years IES Answers

IES-1. Ans. (b)

IES-2. Ans. (d) Intensive property: Whose value is independent of the size or extent i.e. mass of

the system.

Specific property: It is a special case of an intensive property. It is the value of an

extensive property per unit mass of system (Lower case letters as symbols) e.g., specific volume, density (v, ρ).

IES-3. Ans. (a) Extensive property is dependent on mass of system. Thus volume is extensive

property.

IES-4. Ans. (a) Extensive property: Whose value depends on the size or extent i.e. mass of the

system (upper case letters as the symbols) e.g., Volume, Mass (V, M). If mass is increased, the value of extensive property also increases.

Basic Concepts

S K Mondal’s

Chapter 1

• But remember 100% heat can’t be convertible to work but 100% work can be converted to heat. It depends on second law of thermodynamics.

• A thermodynamic system is defined as a definite quantity of matter or a region in space upon which attention is focused in the analysis of a problem.

• The system is a macroscopically identifiable collection of matter on which we focus our attention

IES-6. Ans. (b) In closed thermodynamic system, there is no mass transfer but energy transfer

exists.

IES-7. Ans. (b)

IES-8. Ans. (d) Isentropic means reversible adiabatic. Heat transfer in any finite temp difference is

irreversible.

IES-9. Ans. (a) The energy transfer as heat and work during the forward process as always

identically equal to the energy transfer is heat and work during the reversal or the process is the correct reason for maximum efficiency because it is conservative system.

IES-10. Ans. (b) Isolated System - in which there is no interaction between system and the

surroundings. It is of fixed mass and energy, and hence there is no mass and energy transfer across the system boundary.

IES-11. Ans. (a) All temperature measurements are based on Zeroth law of thermodynamics. IES-12. Ans. (a) Entropy - related to second law of thermodynamics.

Internal Energy (u) = f (T) only (for an ideal gas) Van der Wall's equation related to => real gas.

IES-13. Ans. (d) IES-14. Ans. (d) IES-15.Ans. (d) 0 300 0 150 C 100 300 100 0 C _C − ₌ − _{⇒ = °} − −

IES-16. Ans. (b) Both A and R are correct but R is not correct explanation for A. Temperature is

independent of thermometric property of fluid.

IES-17. Ans. (d)

IES-18. Ans. (d) But it will occur at absolute zero temperature.

IES-19. Ans. (c) Time Constants: The time constant is the amount of time required for a

thermocouple to indicated 63.2% of step change in temperature of a surrounding media. Some of the factors influencing the measured time constant are sheath wall thickness, degree of insulation compaction, and distance of junction from the welded cap on an ungrounded thermocouple. In addition, the velocity of a gas past the thermocouple probe greatly influences the time constant measurement. In general, time constants for measurement of gas can be estimated to be ten times as long as those for measurement of liquid. The time constant also varies inversely proportional to the square root of the velocity of the media.

IES-20. Ans. (c) IES-21. Ans. (a)

IES-22. Ans. (c) In example of (c), it is a case of free expansion heat transfer, work done, and

Basic Concepts

S K Mondal’s

Chapter 1

IES-23. Ans. (a) Work done (W) = P

Δ

V = 100

×

(V2 – V1) = 100

×

2 1 m m ⎛ ⎞ − ⎜_{ρ ρ} ⎟ ⎝ ⎠ = 100 kPa

×

1 1 999 916 ⎛ ₋ ⎞ ⎜ ⎟ ⎝ ⎠ = –9.1 J IES-24. Ans. (d)

W

_A

=

∫

pdV

= × − =

4 (2 1)

4 kJ

1

3 (7 4)

4.5 kJ

2

1 (12 9)

3 kJ

=

= × × −

=

=

= ×

− =

∫

∫

B C

W

pdV

W

pdV

IES-25. Ans. (c) Turbine work = area under curve R–S

(

)

(

)

3 5 1 bar 0.2 0.1 m 1000 Nm 10 0.2 0.1 Nm 1000Nm 11000 Nm pdv = = × − + = × − + =

∫

IES-26. Ans. (b)

IES-27. Ans. (d) Isothermal work is minimum of any process.

0[ is onstant] pv mRT pdv vdp T c pdv vdp = + = = −

∫

∫

∵

IES-28. Ans. (b) For steady flow process, reversible work given by

2

1

vdp

−

_∫

.

IES-29. Ans. (c) A is true because change in heat and work are path functions and thus can't be

expressed simply as difference between the end states. R is false because both work and heat are inexact differentials.

Previous 20-Years IAS Answers

IAS-1. Ans. (c) Intensive properties, i.e. independent of mass are pressure, temperature, velocity

and viscosity. Extensive properties, i.e. dependent on mass of system are volume, electric charge, magnetisation, and potential energy. Thus correct choice is (c).

IAS-2. Ans. (a) IAS-3. Ans. (c)

IAS-4. Ans. (d) The boiling and freezing points on new scale are 400° N and 100°N i.e. range is

300°N corresponding to 100°C. Thus conversion equation is °N = 100 + 3 × °C = 100+ 3 × 60 = 100 + 180 = 280 °N

IAS-5. Ans. (c)

IAS-6. Ans. (c) Since vacuum does not offer any resistance, there is no work transfer involved in

free expansion.

IAS-7. Ans. (b)

IAS-8. Ans. (d) Isothermal work is minimum of any process. IAS-9. Ans. (a)

Basic Concepts

S K Mondal’s

Chapter 1

IAS-12. Ans. (b) Start with D. ∫PdV only valid for quasi-static path so choice (c) & (d) out. Automatically C-4 then eye on A and B. Bottle filling of gas is variable flow so A-2.

IAS-13. Ans. (c) W =

∫

pdv where pressure (p) is an intensive property and volume (v) is an extensive property

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Fir

ondal’s

First

D

O

BJEC

Previou

ion of F

ata for Qu

d the outle n adiabatic cated in t as usually mass flow r e turbine (in 12.157 sume the ab ter at the ergy effects 0.293 e following ermodynam

rst Law

s

t Law

CTIVE

Q

us 20-Y

First Law

uestions Q

et conditio c steam tur the figure. followed. rate of stea n MW) is: bove turbin inlet to the s, the specifi ( g four fig mic cycle, on

w of T

w of T

Q

UEST

Years

w to Ste

Q1 and Q2

ns of rbine The m through (b) 12.941 ne to be par e pump is fic work (in

b) 0.35 1 ures have n the p-v and

Therm

herm

IONS

(G

GATE

eady Flo

2:

the turbin (c rt of a simp 1000 kg/m3_. kJ/kg) supp (c) been dra d T-s planes

odyna

modyn

GATE

E Ques

ow Proc

e is 20 kg/s c) 168.001 ple Rankine . Ignoring k plied to the ) 2.930 awn to re s.

amics

Cha

namic

, IES,

stions

cess S.F

[GA

s the power

[GA

(d) e cycle. The kinetic and pump is: [G (d) 3 epresent a [G

apter 2

cs

IAS)

F.E.E

ATE-2009]

r output of

ATE-2009]

168.785 density of d potential GATE-2009] 3.510 fictitious GATE-2005]

First Law of Thermodynamics

S K Mondal’s

Chapter 2

According to the first law of thermodynamics, equal areas are enclosed by

(a) Figures 1and 2 (b) Figures 1and 3 (c) Figures 1and 4 (d) Figures 2 and 3

Internal Energy – A Property of System

GATE-4. A gas contained in a cylinder is compressed, the work required for compression being 5000 kJ. During the process, heat interaction of 2000 kJ causes the surroundings to the heated. The change in internal energy of the

gas during the process is: [GATE-2004]

(a) – 7000 kJ (b) – 3000 kJ (c) + 3000 kJ (d) + 7000 kJ

Discharging and Charging a Tank

GATE-5. A rigid, insulated tank is initially evacuated. The tank is connected with a supply line through which air (assumed to be ideal gas with constant specific heats) passes at I MPa, 350°C. A valve connected with the supply line is opened and the tank is charged with air until the final pressure inside the tank reaches I MPa. The final temperature inside the tank

(A) Is greater than 350°C (B) Is less than 350°C (C) Is equal to 350°C

(D) May be greater than, less than, or equal to 350°C, depending on the volume of the tank

Previous 20-Years IES Questions

First Law of Thermodynamics

IES-1. Which one of the following sets of thermodynamic laws/relations is directly

involved in determining the final properties during an adiabatic mixing

process? [IES-2000]

(a) The first and second laws of thermodynamics

(b) The second law of thermodynamics and steady flow relations (c) Perfect gas relationship and steady flow relations

(d) The first law of thermodynamics and perfect gas relationship

IES-2. Two blocks which are at different states are brought into contact with each

other and allowed to reach a final state of thermal equilibrium. The final

temperature attained is specified by the [IES-1998]

(a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics

First Law of Thermodynamics

S K Mondal’s

Chapter 2

IES-3. For a closed system, the difference between the heat added to the system and

the work done by the system is equal to the change in [IES-1992]

(a) Enthalpy (b) Entropy

(c) Temperature (d) Internal energy

IES-4. An ideal cycle is shown in the figure. Its

thermal efficiency is given by

3 3 1 1 2 2 1 1

1

1

1

(a)1

(b) 1

1

γ

1

⎛

⎞

⎛

⎞

−

−

⎜

⎟

⎜

⎟

⎝

⎠

⎝

⎠

−

−

⎛

⎞

⎛

⎞

−

−

⎜

⎟

⎜

⎟

⎝

⎠

⎝

⎠

v

v

v

v

p

p

p

p

(

)

(

23 11

)

11

(

(

23 11

)

)

11

1

(c)1

γ

(b) 1

γ

−

−

−

−

−

−

v

v

p

v

v

p

p

p

v

p

p

v

[IES-1998]

IES-5. Which one of the following is correct? [IES-2007]

The cyclic integral of

(

δ

Q

−

δ

W

)

for a process is:

(a) Positive (b) Negative (c) Zero (d) Unpredictable

IES-6. A closed system undergoes a process 1-2 for which the values of Q1-2 and W1-2 are

+20 kJ and +50 kJ, respectively. If the system is returned to state, 1, and Q2-1 is

-10 kJ, what is the value of the work W2-1? [IES-2005]

(a) + 20 kJ (b) –40 kJ (c) –80 kJ (d) +40 kJ

IES-7. A gas is compressed in a cylinder by a movable piston to a volume one-half of

its original volume. During the process, 300 kJ heat left the gas and the

internal energy remained same. What is the work done on the gas? [IES-2005]

(a) 100kNm (b) 150 kNm (c) 200 kNm (d) 300 kNm

IES-8. In a steady-flow adiabatic turbine, the changes in the internal energy,

enthalpy, kinetic energy and potential energy of the working fluid, from inlet to exit, are -100 kJ/kg, -140 kJ/kg, -10 kJ/kg and 0 kJ/kg respectively. Which one of the following gives the amount of work developed by the turbine? [IES-2004]

(a) 100 kJ/kg (b) 110 kJ/kg (c) 140 kJ/kg (d) 150 kJ/kg

IES-9. Gas contained in a closed system consisting of piston cylinder arrangement is

expanded. Work done by the gas during expansion is 50 kJ. Decrease in internal energy of the gas during expansion is 30 kJ. Heat transfer during the

process is equal to: [IES-2003]

(a) –20 kJ (b) +20 kJ (c) –80 kJ (d) +80 kJ

IES-10. A system while undergoing a cycle [IES-2001]

First Law of Thermodynamics

S K Mondal’s

Chapter 2

Process Q kJ/min W kJ/min

A–B B–C C–D D–A +687 -269 -199 +75 +474 0 -180 0 The power developed in kW is, nearly,

(a) 4.9 (b) 24.5 (c) 49 (d) 98

IES-11. The values of heat transfer and work transfer for four processes of a

thermodynamic cycle are given below: [IES-1994]

Process Heat Transfer (kJ) Work Transfer (kJ)

1 2 3 4 300 Zero -100 Zero 300 250 -100 -250 The thermal efficiency and work ratio for the cycle will be respectively.

(a) 33% and 0.66 (b) 66% and 0.36. (c) 36% and 0.66 (d) 33% and 0.36.

IES-12. A tank containing air is stirred by a paddle wheel. The work input to the

paddle wheel is 9000 kJ and the heat transferred to the surroundings from the

tank is 3000 kJ. The external work done by the system is: [IES-1999]

(a) Zero (b) 3000 kJ (c) 6000 kJ (d) 9000 kJ

Internal Energy – A Property of System

IES-13. For a simple closed system of constant composition, the difference between the

net heat and work interactions is identifiable as the change in [IES-2003]

(a) Enthalpy (b) Entropy (c) Flow energy (d) Internal energy

IES-14. Assertion (A): The internal energy depends on the internal state of a body, as

determined by its temperature, pressure and composition. [IES-2006]

Reason (R): Internal energy of a substance does not include any energy that it may possess as a result of its macroscopic position or movement.

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false

(d) A is false but R is true

IES-15. Change in internal energy in a reversible process occurring in a closed system

is equal to the heat transferred if the process occurs at constant: [IES-2005]

(a) Pressure (b) Volume (c) Temperature (d) Enthalpy

IES-16. 170 kJ of heat is supplied to a system at constant volume. Then the system

rejects 180 kJ of heat at constant pressure and 40 kJ of work is done on it. The system is finally brought to its original state by adiabatic process. If the initial value of internal energy is 100 kJ, then which one of the following statements

is correct? [IES-2004]

First Law of Thermodynamics

S K Mondal’s

Chapter 2

(b) The highest value of internal energy occurs at the end of constant pressure process. (c) The highest value of internal energy occurs after adiabatic expansion

(d) Internal energy is equal at all points

IES-17. 85 kJ of heat is supplied to a closed system at constant volume. During the next process, the system rejects 90 kJ of heat at constant pressure while 20 kJ of work is done on it. The system is brought to the original state by an adiabatic process. The initial internal energy is 100 kJ. Then what is the quantity of

work transfer during the process? [IES-2009]

(a) 30 kJ (b) 25 kJ (c) 20 kJ (d) 15 kJ

IES-18. A system undergoes a process during which the heat transfer to the system per

degree increase in temperature is given by the equation: [IES-2004]

dQ/dT = 2 kJ/°C The work done by the system per degree increase in temperature is given by the equation dW/dT = 2 – 0.1 T, where T is in °C. If during the process, the temperature of water varies from 100°C to 150°C, what will be the change in internal energy?

(a) 125 kJ (b) –250 kJ (c) 625 kJ (d) –1250 kJ

IES-19. When a system is taken from state A to

state B along the path A-C-B, 180 kJ of heat flows into the system and it does 130 kJ of work (see figure given):

How much heat will flow into the system along the path A-D-B if the work done by it along the path is 40 kJ?

(a) 40 kJ (b) 60 kJ

(c) 90 kJ (d) 135 kJ [IES-1997]

IES-20. The internal energy of a certain system is a function of temperature alone and

is given by the formula E = 25 + 0.25t kJ. If this system executes a process for which the work done by it per degree temperature increase is 0.75 kJ/K, then

the heat interaction per degree temperature increase, in kJ, is: [IES-1995]

(a) –1.00 (b) –0.50 (c) 0.50 (d ) 1.00

IES-21. When a gas is heated at constant pressure, the percentage of the energy

supplied, which goes as the internal energy of the gas is: [IES-1992]

(a) More for a diatomic gas than for triatomic gas

(b) Same for monatomic, diatomic and triatomic gases but less than 100% (c) 100% for all gases

(d) Less for triatomic gas than for a diatomic gas

Perpetual Motion Machine of the First Kind-PMM1

IES-22. Consider the following statements: [IES-2000]

1. The first law of thermodynamics is a law of conservation of energy.

2. Perpetual motion machine of the first kind converts energy into equivalent work.

First Law of Thermodynamics

S K Mondal’s

Chapter 2

4. The second law of thermodynamics stipulates the law of conservation of energy and entropy.

Which of the statements are correct?

(a) 1 and 2 (b) 2 and 4 (c) 2, 3 and 4 (d) 1, 2 and 3

Enthalpy

IES-23. Assertion (A): If the enthalpy of a closed system decreases by 25 kJ while the

system receives 30 kJ of energy by heat transfer, the work done by the system

is 55 kJ. [IES-2001]

Reason (R): The first law energy balance for a closed system is (notations have their usual meaning)

Δ = −

E

Q W

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false

(d) A is false but R is true

Application of First Law to Steady Flow Process S.F.E.E

IES-24. Which one of the following is the steady flow energy equation for a boiler?

(a) 2 2 1 2 1 2

2

2

v

v

h

h

gJ

gJ

+

=

+

(b)

Q

=

(

h

₂

−

h

₁

)

[IES-2005] (c) 2 2 1 2 1 2

2

2

v

v

h

Q

h

gJ

gJ

+

+ =

+

(d)

W

_s

=

(

h

₂

−

h

₁

)

+

Q

IES-25. A 4 kW, 20 litre water heater is switched on for 10 minutes. The heat capacity

Cp for water is 4 kJ/kg K. Assuming all the electrical energy has gone into

heating the water, what is the increase of the water temperature? [IES-2008]

(a) 15°C (b) 20°C (c) 26°C (d) 30°C

Discharging and Charging a Tank

IES-26. An insulated tank initially contains 0.25 kg of a gas with an internal energy of 200 kJ/kg .Additional gas with an internal energy of 300 kJ/kg and an enthalpy of 400 kJ/kg enters the tank until the total mass of gas contained is 1 kg. What

is the final internal energy(in kJ/kg) of the gas in the tank? [IES-2007]

(a) 250 (b) 275 (c) 350 (d) None of the above

Previous 20-Years IAS Questions

IAS-1. A system executes a cycle during which there are four heat transfers: Q12 = 220

kJ, Q23 = -25kJ, Q34 = -180 kJ, Q41 = 50 kJ. The work during three of the

processes is W12 = 15kJ, W23 = -10 kJ, W34 = 60kJ. The work during the process 4

-1 is: [IAS-2003]

First Law of Thermodynamics

S K Mondal’s

Chapter 2

IAS-2. Two ideal heat engine cycles

are represented in the given figure. Assume VQ = QR, PQ = QS and UP =PR =RT. If the work interaction for the rectangular cycle (WVUR) is 48 Nm, then the work interaction for the other cycle PST is:

(a) 12Nm (b) 18 Nm

(c) 24 Nm (d) 36 Nm IAS-2001]

IAS-3. A reversible heat engine operating between hot and cold reservoirs delivers a work output of 54 kJ while it rejects a heat of 66 kJ. The efficiency of this

engine is: [IAS-1998]

(a) 0.45 (b) 0.66 (c) 0.75 (d) 0.82

IAS-4. If a heat engine gives an output of 3 kW when the input is 10,000 J/s, then the

thermal efficiency of the engine will be: [IAS-1995]

(a) 20% (b) 30% (c) 70% (d) 76.7%

IAS-5. In an adiabatic process, 5000J of work is performed on a system. The system

returns to its original state while 1000J of heat is added. The work done during

the non-adiabatic process is: [IAS-1997]

(a) + 4000J (b) - 4000J (c) + 6000J (d) - 6000J

IAS-6. In a thermodynamic cycle consisting of four processes, the heat and work are

as follows: [IAS-1996]

Q: + 30, - 10, -20, + 5 W: + 3, 10, - 8, 0

The thermal efficiency of the cycle will be:

(a) Zero (b) 7.15% (c) 14.33% (d) 28.6%

IAS-7. Match List-I (Devices) with List-II (Thermodynamic equations) and select the

correct answer using the codes below the lists: [IAS-1996]

List-I List-II A. Turbine 1. W = h2 – h1 B. Nozzle 2. h1 = h2 C. Valve 3. h1 = h2 + V2/2 D. Compressor 4. W = h1 – h2 Codes: A B C D A B C D (a) 4 3 2 1 (b) 2 3 1 4 (c) 4 3 1 2 (d) 3 2 4 1

First Law of Thermodynamics

S K Mondal’s

Chapter 2

IAS-8. Given that the path 1-2-3, a system

absorbs 100kJ as heat and does 60kJ work while along the path 1-4-3 it does 20kJ work (see figure given). The heat absorbed during the cycle 1-4-3 is:

(a) - 140 Kj (b) - 80 kJ (c) - 40kJ (d) + 60 kJ

[IAS 1994] IAS-9. The given figure shows the

variation of force in an elementary system which undergoes a process during which the plunger position changes from 0 to 3 m. lf the internal energy of the system at the end of the process is 2.5 J higher, then the heat absorbed

during the process is: [IAS-1994]

(a) 15 J (b) 20 J (c) 25 J (d) 30 J

IAS-10. The efficiency of a reversible cyclic process undergone by a substance as shown in the given diagram is:

(a) 0.40 (b) 0.55 (c) 0.60 (d) 0.80

[IAS 1994]

Internal Energy – A Property of System

IAS-11. Which one of the following is the correct expression for change in the internal

energy for a small temperature change

Δ

T for an ideal gas? [IAS-2007]

(a)

Δ =

U

C

_v

× Δ

T

(b)

Δ =

U

C

_p

× Δ

T

(c) p v

C

U

T

C

Δ =

× Δ

(d)

Δ =

U

(

C

_p

−

C

_v

)

× Δ

T

IAS-12. The heat transferred in a thermodynamic cycle of a system consisting of four

processes is successively 0, 8, 6 and -4 units. The net change in the internal

energy of the system will be: [IAS-1999]

(a) – 8 (b) Zero (c) 10 (d) –10

IAS-13. During a process with heat and work interactions, the internal energy of a

system increases by 30 kJ. The amounts of heat and work interactions are

First Law of Thermodynamics

S K Mondal’s

Chapter 2

(a) - 50 kJ and - 80 kJ (b) -50 kJ and 80 kJ (c) 50 kJ and 80 kJ (d) 50 kJ and - 80 kJ

IAS-14. A mixture of gases expands from 0.03 m3 _{to 0.06 m}3 _{at a constant pressure of 1}

MPa and absorbs 84 kJ of heat during the process. The change in internal

energy of the mixture is: [IAS 1994]

(a) 30 kJ (b) 54 kJ (c) 84 kJ (d) 114 kJ

IAS-15. In an adiabatic process 6000 J of work is performed on a system. In the

non-adiabatic process by which the system returns to its original state 1000J of heat is added to the system. What is the work done during non-adiabatic

process? [IAS-2004]

(a) + 7000 J (b) - 7000 J (c) + 5000 J (d) - 5000 J

Enthalpy

IAS-16. The fundamental unit of enthalpy is: [IAS 1994]

(a) MLT-2 _{(b) ML}-2_T-1 _{(c) ML}2_T-2 _{(d) ML}3_T-2

Application of First Law to Steady Flow Process S.F.E.E

IAS-17. In a test of a water-jacketed compressor, the shaft work required is 90 kN-m/kg of air compressed. During compression, increase in enthalpy of air is 30 kJ/kg of air and increase in enthalpy of circulating cooling water is 40 kJ/ kg of air. The change is velocity is negligible. The amount of heat lost to the atmosphere

from the compressor per kg of air is: [IAS-2000]

(a) 20kJ (b) 60kJ (c) 80 kJ (d) 120kJ

IAS-18. When air is compressed, the enthalpy is increased from 100 to 200 kJ/kg. Heat

lost during this compression is 50 kJ/kg. Neglecting kinetic and potential energies, the power required for a mass flow of 2 kg/s of air through the

compressor will be: [IAS-1997]

(a) 300 kW (b) 200 kW (c) 100 kW (d) 50 kW

Variable Flow Processes

IAS-19. Match List-I with List-II and select the correct answer using the codes given

below Lists: [IAS-2004]

List-I List-II

A. Bottle filling of gas 1. Absolute zero temperature

B. Nernst Simon statement 2. Variable flow

C. Joule Thomson effect 3. Quasistatic path

D.

∫

pdv

4. Isenthalpic process

5. Dissipative effect 6. Low grade energy

First Law of Thermodynamics

S K Mondal’s

Chapter 2

7. Process and temperature during phase

change

Codes: A B C D A B C D

(a) 6 5 4 3 (b) 2 1 4 3 (c) 2 5 7 4 (d) 6 1 7 4

IAS-20. A gas chamber is divided into two parts by means of a partition wall. On one

side, nitrogen gas at 2 bar pressure and 20°C is present. On the other side, nitrogen gas at 3.5 bar pressure and 35°C is present. The chamber is rigid and thermally insulated from the surroundings. Now, if the partition is removed,

(a) High pressure nitrogen will get throttled [IAS-1997]

(b) Mechanical work, will be done at the expense of internal energy (c) Work will be done on low pressure nitrogen

First Law of Thermodynamics

S K Mondal’s

Chapter 2

Answers with Explanation (Objective)

Previous 20-Years GATE Answers

GATE-1. Ans. (a)

+

+

+

=

+

+

+

2 2 1 1 1 2 1 2

gZ

dQ

gZ

dW

h

h

2000 1000

dm

2000 1000

dm

C

C

×

×

+

+

=

+

+

+

+

+

= +

2 2

160

9.81 10

100

9.81 6 dW

3200

2600

2000

1000

2000

1000

dm

dW

600 7.8 0.04

dm

GATE-2. Ans. (c)

(

)

(

)

ν

=

₂

−

₁

=

1

−

×

W

P

P

3000 70

kJ/kg = 2.93

1000

GATE-3. Ans. (a) Fig-1 & 2 both are power cycle, so equal areas but fig-3 & 4 are reverse power

cycle, so area is not meant something.

GATE-4. Ans. (c) 2 1 2 1 2 1 dQ du dw Q u u W or 2000 u u 5000 or u u 3000kJ = + = − + − = − − − =

GATE-5. Ans (a) The final Temp. (T2)=

γ

T

₁

Previous 20-Years IES Answers

IES-1. Ans. (a) If we adiabatically mix two liquid then perfect gas law is not necessary. But entropy

change in the universe must be calculated by Second law of thermodynamics. Final entropy of then system is also a property. That so why we need second law.

IES-2. Ans. (b) Using conservation of energy law we may find final temperature.

IES-3. Ans. (d) From First law of thermodynamics, for a closed system the net energy transferred

as heat Q and as work W is equal to the change in internal energy, U, i.e. Q – W = U

IES-4. Ans. (c) Total heat addition is constant volume heat addition,Q₁₂=c T_v( ₂−T₁)

Total heat rejection is constant pressure heat rejection, Q₃₁=c T_p( ₃ −T₁)

Now from equation of state

1 2 2 2 1 1 2 1 3 3 1 3 1 1 3 1 ( const.) ( const.) P P P v or T T T T P v v v and p or T T T T v = = = × = = = × ∵ ∵ Efficiency, 31 3 1 3 1 12 2 1 2 1 ( ) ( ) 1 1 1 ( ) ( ) p v c T T Q T T Q c T T T T

η

= − = − − = −

γ

− − −

S

IE IE IE IE IE IE IE IE IE IE

S K Mo

or S-5. Ans. (c zero S-6. Ans. (b or S-7. Ans. (d The S-8. Ans. (d Q O or W − − Cha S-9. Ans. (b

Δ

E

Δ

W Q = S-10. Ans. (a S-11. Ans. (b

Wo

S-12. Ans. pro per can volu wor inte S-13. Ans. (d S-14. Ans. ( Inte resu the

Fir

ondal’s

⎛ ⎜ ⎝ = − ⎛ ⎜ ⎝

η

γ

2 1 v v P P c) It is du = đ o. ) ΣdQ= ΣdW

(

)

+ − = 20 10 ) dQ = du erefore du = 0 ) x x x W h W 140 W 150 kJ / ⎛ − = Δ_⎜ + ⎝ − = − − = ange of inter ) Q =

Δ

E+

Δ

E = –30 kJ (d W = + 50 kJ ( = –30 + 50 = + a) Net work b)

Wo

hea

th

η

=

ork ratio

=

∑

(a) This is cess or an rforming wor n be raised. ume process rk on the sur ernal energy d)

(a) The inter

ernal energy ult of its ma re.

rst Law

s

⎞ × − ⎟ ⎠ = ⎞ × _{− ⎟} ⎠ 3 1 1 1 2 1 1 1 v T T v P T T P đQ – đW, as u W − + _{2 1} 50 W + dw as u = 0 or dQ = d 2 V gz 2 10 0 / kg ⎞ + _⎟ ⎠ − + nal energy =

Δ

W decrease in in (work done b + 20 kJ =

∑

dW=47

rk done

30

at added

=

( )

( )

w

w

+

−

+

∑

∑

∑

s a case of n is isocho rk on the sys In an irrev , the system rrounding at . rnal energy d y of a substa acroscopic

w of T

− = − −

γ

3 2 ( 1 ( v v p p u is a thermo − 1 2 or Q − = 2 1 or W = const. dw = 300kNm -100 kJ/kg i nternal energ y the system 74 – 180 kJ/m

00 100

0.

300

−

₌

( )

550

5

w

−

=

constant vo oric process. stem tempera versible con m doesn't per the expense depends only ance does no position or m

Therm

1 1 1 1 ) ) v p p v odynamic pro − − + = 2 Q2 1 W1 = −40kJ m s superfluou gy) m) min = 294 kJ

.66

350

0.36

550

−

₌

olume . By ature stant rform of its y upon the in ot include an movement. T

odyna

operty and it −2+W2 1− s data. J/min = 294/6

6

nitial and fin ny energy th That so why

amics

Cha

ts cyclic integ 60 kJ/s = 4.9 nal states of hat it may p in SFEE v2_/

apter 2

gral must be kW the system. possess as a /2 and gz is

S

IE IE IE IE IE IE

S K Mo

If in Bu Mic par and gas S-15. Ans. (b S-16. Ans. (a S-17. Ans. (d S-18. Ans. (c dQ 2.d or S-19. Ans. ( be s S-20. Ans. (d

Fir

ondal’s

nternal energ r Remembe croscopic vie rticle consist d specific el s, form the sp b) dQ dU= + a) d) c)

(

du dw t du 2 0 du 0.1Td = + = + − =

∫

∫

(c) Change o same even al d) dQ=dE

rst Law

s

gy include po er: w of a gas s of transla lectronic en pecific interna pdV if V +

)

2 0.1T dT 0.1 dT T 2 ⎡ = _{× ⎣} f internal en long path AD dW +

w of T

osition or mo is a collectio ational ene nergy. All th al energy, e , V is cons tan t 150 100 0.1 ₁₅₀ 2 ⎤ = ⎡ ⎦ ⎣ nergy from A DB. ∴ Heat fl d or d

Therm

vement then on of particl rgy, rotatio hese energies , of the gas.

( ) (

dQ v = 2 Q =180 1 3 U 10 U 27 ∴ = = For the p For the p For the p For a cyc ⇒ 85-90+ ⇒ ⇒ 2 2 0 −100 ⎤_⎦=6 to B along p flow along AD dQ dE d dt = dt + d

odyna

n why this v2_/ les in rando onal energy summed ove

( )

dU v 0kJ= Δ + Δu W 2 0kJ, U 10 0 180 40 = − + = process 1-2 dQ = process 2-3 dQ = process 3-1 dQ = clic process ∑ dQ +0 = 0-20+dW -5 = -2 dW= -20+5 625kJ path ACB = 1 DB = 40 + 50 dW dt

amics

Cha

/2 and gz term om motion. E y, vibration

er all the par

W= Δ + −u ( 40) 0 170 270 130 kJ + = = = +85 ,dW = 0 = -90 kJ, dW = 0, dW = ? = ∑ dW W 0+dW 5 = +15kJ 180 - 130 = 5 = 90 kJ.

apter 2

ms is there. Energy of a nal energy rticles of the ) 0 kJ, 0 = -20kJ 50 kJ. It will

First Law of Thermodynamics

S K Mondal’s

Chapter 2

Given: E 25 0.25t kJ and 0.75 / then 0.25 / Therefore 0.25 0.75 / 1.00 / dW kJ k dt dE kJ K dt dQ dE dW kJ K kJ K dt dt dt = + = = = + = + =

IES-21. Ans. (a)

IES-22. Ans. (a) A closed system does exchange work or energy with its surroundings. option ‘3’ is

wrong. 4. “The law of conservation of entropy” is imaginary so option ‘4’ is also wrong.

IES-23. Ans. (a)

IES-24. Ans. (b) 12 22 1 1 2 2 v dQ v dw h gz h gz 0 2 dm 2 dm + + + = + + + =

For boiler v1, v2 is negligible and z1 = z2 and dw 0

dm = or dQ

(

h₂ h₁

)

dm = − IES-25. Ans. (d)

(

)

P mC T = 4 10 60 20 4 T = 2400 T = 30C Δ × × ⇒ × × Δ ⇒ Δ °

IES-26. Ans. (c) Enthalpy of additional gas will be converted to internal energy.

Uf= miui+(mf-mi)hp = 0.25x200+(1-0.25)x400 = 350 kJ

As total mass = 1kg, uf=350 kJ/kg

Note: You cannot simply use adiabatic mixing law here because it is not closed system.

This is a problem of variable flow process. If you calculate in following way it will be wrong.

Final internal energy of gas(mixture) is

u = 1 1 2 2 1 2 m u m u m m + + u = kJ kJ (0.25kg) 200 (0.75kg) 300 kg kg (0.25 0.75)kG ⎛ ⎞₊ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + u = 275 kJ kg

It is valid for closed system only.

Previous 20-Years IAS Answers

IAS-1. Ans. (b)

∑

dQ=

∑

dW or 220 -25 -180 +50 = 15 -10 +60 +W4-1

First Law of Thermodynamics

S K Mondal’s

Chapter 2

Areas

Δ

PTS=

1

2

Area (WVUR)

∴

Work PTS=

1

48

2

×

=24 Nm

IAS-3. Ans. (a) work output work out put 54 0.45

Heat input work output heat rejection 54 66

η = = = = + + IAS-4. Ans. (b) 3 3 10 watts Thermal efficiency 0.3 30% 10.000 J/s W Q × = = = = IAS-5. Ans. (c)

(

)

(

) (

)

(

)

(

)

(

)

(

)

1 2 2 1 1 2 2 1 2 1 2 1 1 2 2 1 2 1 2 1 1 2 2 1 2 1 Q U U W or 0 U U 5000 or U U 5000J Q U U W or W Q U U Q U U 1000 5000 6000J − − − − − − − = − + = − + − − = = − + = − − = + − = + =

IAS-6. Ans. (c) Net work output = 3 + 10 – 8 = 5 unit and Heat added = 30 + 5 = 35 unit Therefore efficiency, 5 100% 14.33%

35

η= × =

IAS-7. Ans. (a)

IAS-8. Ans. (d) Q123 = U13 + W123 or, 100 = U13 + 60 or, U13 = 40 kJ

And Q143 = U13 + W143 = 40+20 = 60 kJ

IAS-9.Ans. (b) Total work = 5 × 3 +

5

1

17

.

5

J

2

1

_×

_×

₌

or

δ

W

=

du

+

δ

W

=

2

.

5

+

17

.

5

=

20

J

IAS-10. Ans. (c) Efficiency Area under 500 and 1500

Area under 0 and 1500

=

{

}

{

}

1 _{(5 1) (4 2) (1500 500)} 3000 2 _0.6 1 _{(5 1) (4 2) (1500 500) (5 1) 500} 5000 2 × − + − × − = = = × − + − × − + − ×

IAS-11. Ans. (a)

IAS-12. Ans. (b) Internal energy is a property of a system so

_∫

du 0=

IAS-13. Ans. (a) dQ du dW if du= + = +30kJ then dQ= −50kJ and dW= −80kJ

IAS-14. Ans. (b)

δ

W

=

du

+

δ

W

=

du

+

pdV

or 84 × 103_{J = du + 1 × 10}6 _{× (0.06 – 0.03) = du +30 kJ or du = 83 – 30 = 54 kJ}

IAS-15. Ans. (a) Q1-2 = U2 –U1 +W1-2

Or 0 = U2 –U1 - 6000 or U2 –U1 = +6000 Q2-1 = U1-U2+W2-1 or W2-1 = Q2-1 - (U1-U2) =1000+6000=7000J IAS-16. Ans. (c)

First Law of Thermodynamics

S K Mondal’s

Chapter 2

IAS-17. Ans. (a) Energy balance gives as

( )

( )

= Δ + Δ + = − − = air water dW _{h h} dQ dm dm dQ or 90 30 40 dm 20kJ / kg of air compressed.

IAS-18. Ans. (a)

( )

( )

(

)

(

)

1 2 1 2 dQ dw m h m h dt dt dw dQ or m h h 2 100 200 50 2 300kW dt dt

i.e. 300kW work have to given to the system.

+ = +

= − + = × − − × = −

IAS-19. Ans. (b) IAS-20. Ans. (a)